How do you find the important points to graph f(x)=2x^2?

1 Answer
Tony B
Jul 10, 2018

See explanation

Explanation:

Set y=f(x)=2x^2

Compare to the standardised form: y=ax^2+bx+c
This equation color(white)("dddddfdd") ->color(white)("ddddd")y=2x^2+0x+0

Key points:

a>0 -> ("positive") so the graph is of form uu

c-> y-intercept =0

x_("vertex") -> (-1/2)xx b/acolor(white)("ddd") =color(white)("ddd") (-1/2)xx0/2color(white)("d")=color(white)("d")0
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

As x_("vertex")=0 the axis of symmetry is the y-axis

As a>0 then the general shape is uu with the y-axis in the middle.

Couple this with c=0 and it means that the vertex is at (x,y)=(0.0)

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